$\begingroup$You want $|2 + x| < 1.$ You can solve for $x$ now. You aren't totally wrong, I am guessing you wanted to take the limit from what you have gotten, but I honestly don't find that limit to be very appetizing.$\endgroup$
– IAmNoOneAug 16 '14 at 23:15

There's an easy formula for geometric series that says it converges if and only if the ratio $r$ has absolute value less than one. In this case you want $-1 < 2 + x < 1$, i.e. $-3 < x < -1$.

However, your way works too. Starting where you left off, we have (for $x \ne -1$)
$$
S_n
= \frac{(2+x)(1-(x+2)^n)}{1-(x+2)}
= \frac{-(2+x)}{1+x}\left(1-(x+2)^n\right)
$$
If $|x + 2| < 1$, then $|x+2|^n \to 0$, so the partial sums converge to $\dfrac{-(2+x)}{1+x}$.
If $|x + 2| > 1$, $|x + 2|^n \to \infty$ and $(2 + x) \ne 0$, so the partial sums diverge. In the last case $|x + 2| = 1$, we have $x = -1$ or $x = -3$, and you can check that both of these give divergent series. For $x = -1$ your partial sum formula doesn't work, so you need to look at the series directly.